Let we have a function \[f\left( x \right) = \sqrt {\cot \left( {5 + 3x} \right)\left( {\cot \left( 5 \right) + \cot \left( {3x} \right)} \right) - \sqrt {\cot 3x} + 1} \], find the domain of the function.
A). \[R - \left\{ {\dfrac{{n\pi }}{3}} \right\},n \in I\]
B). \[\left( {2n + 1} \right)\dfrac{\pi }{6},n \in I\]
C). \[R - \left\{ {\dfrac{{n\pi }}{3},\dfrac{{n\pi - 5}}{3}} \right\},n \in I\]
D). \[R - \left\{ {\dfrac{{n\pi - 5}}{3}} \right\},n \in I\]

Hint: Here, in the question, we are given a function in the form of \[f\left( x \right)\] and asked to find the domain of the function. The domain refers to the set of all possible values of \[x\] for which a function is defined. To find the domain, we will first simplify the given function to a possible extent and then check for the possible input values of \[x\] to get the desired result.
Formula used:
\[\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot B + \cot A}}\]

Complete step-by-step solution:
Given, \[f\left( x \right) = \sqrt {\cot \left( {5 + 3x} \right)\left( {\cot \left( 5 \right) + \cot \left( {3x} \right)} \right) - \sqrt {\cot 3x} + 1} \]
To simplify, we will use trigonometric identity, \[\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot B + \cot A}}\]
Therefore, \[f\left( x \right) = \sqrt {\dfrac{{\cot 5\cot 3x - 1}}{{\cot 3x + \cot 5}}\left( {\cot \left( 5 \right) + \cot \left( {3x} \right)} \right) - \sqrt {\cot 3x} + 1} \]
Now, we can cancel the similar terms in numerator and denominator,
\[\therefore f\left( x \right) = \sqrt {\cot 5\cot 3x - 1 - \sqrt {\cot 3x} + 1} \]
Adding \[1\] and \[ - 1\], we get,
\[\therefore f\left( x \right) = \sqrt {\cot 5\cot 3x - \sqrt {\cot 3x} } \]
To find the domain, Put \[f\left( x \right) = 0\]
\[\therefore \sqrt {\cot 5\cot 3x - \sqrt {\cot 3x} } = 0\]\[\]
Squaring both sides, we get,
\[\cot 5\cot 3x - \sqrt {\cot 3x} = 0\]
Now, the above equation cannot be simplified further. But it is clearly visible that \[x\] is the only independent variable and \[3x\] is the only angle left with its cotangent. Now, we have to find those values for \[x\] such that \[\cot 3x\] is defined. Or, in other words, domain will be all the real numbers except for the numbers where \[\cot 3x\] is undefined. And we know, the cotangent of any angle is undefined at \[n\pi ,n \in I\]. It means \[3x\] should not be equal to \[n\pi ,n \in I\].
Hence, the domain for \[f\left( x \right) = \sqrt {\cot \left( {5 + 3x} \right)\left( {\cot \left( 5 \right) + \cot \left( {3x} \right)} \right) - \sqrt {\cot 3x} + 1} \] is \[R - \left\{ {\dfrac{{n\pi }}{3}} \right\},n \in I\]
Hence the correct option is A. \[R - \left\{ {\dfrac{{n\pi }}{3}} \right\},n \in I\] is the correct option.

Note: Generally, we determine the domain of the function by looking for those values of the independent variable (usually\[x\]) which we are permitted to use. There are two key-points which should be kept in mind while finding the domain of any function: (i) There should not be negative values under a square root sign, and, (ii) There should not be zero in the denominator of the function.